$\int \min \{\frac1{(1+x^2)^2},\frac1{5-x^2}\}$ solution verification

Question

$$\int \min \{\frac1{(1+x^2)^2},\frac1{5-x^2}\}$$

I see this integral as the integral of the function which smaller of these two

$$\int\frac1{(1+x^2)^2}=\frac12\arctan(x)+\frac x{2(1+x^2)}+C_1$$

$$\int\frac1{5-x^2}=\frac{\sqrt5}{10}\log|\sqrt5+x| -\frac{\sqrt5}{10}\log|\sqrt5-x| +C_2$$

$$ \int \min \{\frac1{(1+x^2)^2},\frac1{5-x^2}\}= \cases{ \frac12\arctan(x)+\frac x{2(1+x^2)}+C_1 & $\frac1{(1+x^2)^2}< \frac1{5-x^2}$ \cr \frac{\sqrt5}{10}\log|\sqrt5+x| -\frac{\sqrt5}{10}\log|\sqrt5-x| +C_2 & $\frac1{5-x^2}< \frac1{(1+x^2)^2}$ } $$

Is the idea correct and what happens in points where $\frac1{(1+x^2)^2}=\frac1{5-x^2}$ ?

Answer 1

For $x^2>5$, the function reduces to $\dfrac1{5-x^2}$ and we can use an indefinite integral

$$F(x)=\int\frac{dx}{5-x^2}+C_0.$$ Notice that you may not cross the borders $x^2=5$ and there can be two distinct constants on either sides.

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For $x^2<5$, we first restrict the study to $x\ge0$ for convenience. We switch from one function to the other at $x=1$. So for $x\le1$, we adopt $$F(x)=\int_0^x\frac{dt}{5-t^2}+C_1$$ and for $$1\le x\le \sqrt5,$$

$$F(x)=\int_0^1\frac{dt}{5-t^2}+\int_1^x\frac{dt}{(1+t^2)^2}+C_1=\frac1{\sqrt 5}\text{artanh}\frac1{\sqrt 5}+\int_1^x\frac{dt}{(1+t^2)^2}+C_1.$$

Now for the negative domain, we can use by symmetry

$$F(x)= C_1-F(-x).$$

Answer 2

What did you do to determine which values of x make one smaller than the other? First notice that $\frac{1}{(1+ x^2)^2}$ is always positive while $\frac{1}{5- x^2}$ is positive only for $-\sqrt{5}< x< \sqrt{5}$.

For all $x< -\sqrt{5}$ and all $x> \sqrt{5}$, the minimum if $\frac{1}{5- x^2}$. For $-\sqrt{5}< x< \sqrt{5}$, $\frac{1}{(1+ x^2)^2}< \frac{1}{5- x^2}$ if and only if $5- x^2< (1- x^2)^2= 1- 2x^2+ x^4$.

So we want to find x such that $x^4- x^2- 4> 0$. Let $y= x^2$ and solve $y^2- y- 4> 0$. $y^2- y- 4= y^2- y+ \frac{1}{4}- \frac{1}{4}- 4= (y- \frac{1}{2})^2- \frac{17}{4}> 0$. $(y- \frac{1}{2})^2> \frac{17}{4}$.

That will be true for $y= x^2< \frac{1- \sqrt{17}}{2}$ and for $y= x^2> \frac{1+ \sqrt{17}}{2}$. Of course, for real x, $x^2$ is not negative so we are left with $x^2> \frac{1+ \sqrt{17}}{2}$ which means $x< -\sqrt{\frac{1+ \sqrt{17}}{2}}$ and $x> \sqrt{\frac{1+ \sqrt{17}}{2}}$.